CBSE Class 11 Mathematics Limits And Derivatives Worksheet Set 09

CBSE Class 11 Mathematics Worksheet - Limits and Derivatives. The questions in the worksheets have been specifically designed by best teachers so that the students can practise them to clear their concepts and get better marks in tests and examinations. Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Evaluate \( \lim_{x\to0} \left( \frac{10^x - 2^x - 5^x + 1}{x \tan x} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{10^x - 2^x - 5^x + 1}{x \tan x} \right) \)
\( = \lim_{x\to0} \left( \frac{2^x(5^x-1) - 1(5^x-1)}{x \tan x} \right) \) \quad \( \{10^x = 2^x \cdot 5^x\} \)
\( = \lim_{x\to0} \left( \frac{(5^x-1)(2^x-1)}{x \tan x} \right) \)
\( = \lim_{x\to0} \left( \frac{\left( \frac{5^x-1}{x} \right) \times x \cdot \left( \frac{2^x-1}{x} \right) \times x}{x \left( \frac{\tan x}{x} \right) \times x} \right) \)
\( = \lim_{x\to0} \left( \frac{\left( \frac{5^x-1}{x} \right) \cdot \left( \frac{2^x-1}{x} \right)}{\frac{\tan x}{x}} \right) \)
\( = \frac{\lim_{x\to0} \left( \frac{5^x-1}{x} \right) \times \lim_{x\to0} \left( \frac{2^x-1}{x} \right)}{\lim_{x\to0} \left( \frac{\tan x}{x} \right)} \)
\( = \frac{(\log 5)(\log 2)}{1} \) \quad \( \left\{ \lim_{x\to0} \left( \frac{\tan x}{x} \right) = 1, \lim_{x\to0} \left( \frac{a^x-1}{x} \right) = \log a \right\} \)
\( = \log(5) \log(2) \) ans.

Question. Evaluate \( \lim_{x\to0} \left( \frac{e^x + e^{-x} - 2}{x^2} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{e^x + e^{-x} - 2}{x^2} \right) \)
\( = \lim_{x\to0} \left( \frac{e^x + \frac{1}{e^x} - 2}{x^2} \right) \)
\( = \lim_{x\to0} \left( \frac{e^{2x} + 1 - 2e^x}{e^x \cdot x^2} \right) \) \quad \( \{(e^x)^2 = e^{2x}\} \)
\( = \lim_{x\to0} \left( \frac{(e^x-1)^2}{e^x \cdot x^2} \right) \)
\( = \lim_{x\to0} \left( \frac{e^x-1}{x} \right)^2 \times \lim_{x\to0} \left( \frac{1}{e^x} \right) \)
\( = (1)^2 \times \frac{1}{e^0} \) \quad \( \left\{\lim_{x\to0} \left( \frac{e^x-1}{x} \right) = 1\right\} \)
\( = 1 \) ans. \quad \( \{e^0 = 1\} \)

Question. Evaluate \( \lim_{x\to0} \left( \frac{\log(1+x^3)}{\sin^3 x} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{\log(1+x^3)}{\sin^3 x} \right) \)
\( = \lim_{x\to0} \left[ \frac{\frac{\log(1+x^3)}{x^3} \times x^3}{\frac{\sin^3 x}{x^3} \times x^3} \right] \)
\( = \frac{\lim_{x\to0} \left( \frac{\log(1+x^3)}{x^3} \right)}{\lim_{x\to0} \left( \frac{\sin^3 x}{x^3} \right)} \) \quad \( \left\{\lim_{x\to0} \left( \frac{\log(1+x)}{x} \right) = 1\right\} \)
\( = \frac{1}{1^3} = 1 \) ans.

Question. Evaluate \( \lim_{x\to0} \left( \frac{2^{3x} - 3^{2x}}{\sin(3x)} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{2^{3x} - 3^{2x}}{\sin(3x)} \right) \)
\( = \lim_{x\to0} \left( \frac{2^{3x} - 3^{2x} - 1 + 1}{\sin(3x)} \right) \)
\( = \lim_{x\to0} \left( \frac{(2^{3x}-1) - (3^{2x}-1)}{\sin(3x)} \right) \)
\( = \lim_{x\to0} \left( \frac{ \frac{2^{3x}-1}{3x} \times 3x - \frac{3^{2x}-1}{2x} \times 2x }{\frac{\sin(3x)}{3x} \times 3x} \right) \)
\( = \frac{\lim_{x\to0} \left[ \frac{2^{3x}-1}{3x} \right] \times 3 - \lim_{x\to0} \left[ \frac{3^{2x}-1}{2x} \right] \times 2}{\lim_{x\to0} \left[ \frac{\sin(3x)}{3x} \right] \times 3} \)
\( = \frac{3(\log 2) - 2(\log 3)}{1 \times 3} \) \quad \( \left\{\lim_{x\to0} \left( \frac{a^x-1}{x} \right) = \log a\right\} \)
\( = \frac{\log 2^3 - \log 3^2}{3} \) \quad \( \{\log m^n = n \log m\} \)
\( = \frac{1}{3} \log \left( \frac{8}{9} \right) \) ans. \quad \( \{\log A - \log B = \log \left( \frac{A}{B} \right)\} \)

Question. Evaluate \( \lim_{x\to0} \left( \frac{x(e^x-1)}{1-\cos x} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{x(e^x-1)}{1-\cos x} \right) \)
\( = \lim_{x\to0} \left( \frac{x(e^x-1)}{2 \sin^2 \frac{x}{2}} \right) \)
\( = \lim_{x\to0} \left( \frac{x \cdot \frac{e^x-1}{x} \times x}{2 \cdot \frac{\sin^2 \frac{x}{2}}{\frac{x^2}{4}} \times \frac{x^2}{4}} \right) \)
\( = \frac{2 \lim_{x\to0} \left( \frac{e^x-1}{x} \right)}{\lim_{x\to0} \left( \frac{\sin^2 \frac{x}{2}}{\frac{x^2}{4}} \right)} \) \quad \( \left\{\lim_{x\to0} \left( \frac{e^x-1}{x} \right) = 1\right\} \)
\( = \frac{2(1)}{1^2} = 2 \) ans.

Question. Evaluate \( \lim_{x\to0} \left( \frac{a^x + b^x + c^x - 1}{x} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{a^x + b^x + c^x - 1}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{a^x + b^x + c^x - 1 - 1 - 1}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{(a^x-1) + (b^x-1) + (c^x-1)}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{a^x-1}{x} + \frac{b^x-1}{x} + \frac{c^x-1}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{a^x-1}{x} \right) + \lim_{x\to0} \left( \frac{b^x-1}{x} \right) + \lim_{x\to0} \left( \frac{c^x-1}{x} \right) \)
\( = \log a + \log b + \log c = \log(abc) \) ans. \quad \( \{\log A + \log B = \log(AB)\} \)

Question. Evaluate \( \lim_{x\to0} \left( \frac{a^x + b^x - c^x - d^x}{x} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{a^x + b^x - c^x - d^x}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{a^x + b^x - c^x - d^x - 1 - 1 + 1 + 1}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{(a^x-1) + (b^x-1) - (c^x-1) - (d^x-1)}{x} \right) \)
\( = \lim_{x\to0} \left[ \left( \frac{a^x-1}{x} \right) + \left( \frac{b^x-1}{x} \right) - \left( \frac{c^x-1}{x} \right) - \left( \frac{d^x-1}{x} \right) \right] \)
\( = \lim_{x\to0} \left( \frac{a^x-1}{x} \right) + \lim_{x\to0} \left( \frac{b^x-1}{x} \right) - \lim_{x\to0} \left( \frac{c^x-1}{x} \right) - \lim_{x\to0} \left( \frac{d^x-1}{x} \right) \)
\( = \log a + \log b - \log c - \log d \) \quad \( \left\{\lim_{x\to0} \left( \frac{a^x-1}{x} \right) = \log a\right\} \)
\( = (\log a + \log b) - (\log c + \log d) \)
\( = \log(ab) - \log(cd) \) \quad \( \{\log A + \log B = \log(AB)\} \)
\( = \log \left( \frac{ab}{cd} \right) \) ans. \quad \( \left\{\log A - \log B = \log \left( \frac{A}{B} \right)\right\} \)

Question. Evaluate \( \lim_{x\to0} \left( \frac{\log(5+x) - \log(5-x)}{x} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{\log(5+x) - \log(5-x)}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{\log\left( 5 \left( 1 + \frac{x}{5} \right) \right) - \log\left( 5 \left( 1 - \frac{x}{5} \right) \right)}{x} \right) \)
\( = \lim_{x\to0} \left[ \frac{\left\{ \log 5 + \log \left( 1 + \frac{x}{5} \right) \right\} - \left\{ \log 5 + \log \left( 1 - \frac{x}{5} \right) \right\}}{x} \right] \) \quad \( \{\log A + \log B = \log(AB)\} \)
\( = \lim_{x\to0} \frac{\log\left( 1 + \frac{x}{5} \right) - \log\left( 1 - \frac{x}{5} \right)}{x} \)
\( = \lim_{x\to0} \frac{\log\left( 1 + \frac{x}{5} \right)}{\frac{x}{5} \times 5} - \lim_{x\to0} \frac{\log\left( 1 + \frac{-x}{5} \right)}{x} \)
\( = \lim_{x\to0} \frac{\log\left( 1 + \frac{x}{5} \right)}{\frac{x}{5} \times 5} + \lim_{x\to0} \frac{\log\left( 1 + \frac{-x}{5} \right)}{- \frac{x}{5} \times 5} \)
\( = \frac{1}{5} + \frac{1}{5} \) \quad \( \left\{\lim_{x\to0} \left( \frac{\log(1+x)}{x} \right) = 1\right\} \)
\( = \frac{2}{5} \) ans.

Question. Evaluate \( \lim_{x\to0} \left( \frac{e^{3+x} - \sin x - e^3}{x} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{e^{3+x} - \sin x - e^3}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{e^{3+x} - e^3 - \sin x}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{e^3(e^x-1) - \sin x}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{e^3(e^x-1)}{x} - \frac{\sin x}{x} \right) \)
\( = e^3 \lim_{x\to0} \left[ \frac{e^x-1}{x} \right] - \lim_{x\to0} \left( \frac{\sin x}{x} \right) \)
\( = e^3 - (1) \)
\( = e^3 - 1 \) ans.

Question. Evaluate \( \lim_{x\to0} \left( \frac{9^x - 6^x - 6^x + 4^x}{x} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{9^x - 6^x - 6^x + 4^x}{x^2} \right) \)
\( = \lim_{x\to0} \left( \frac{3^x(3^x-2^x) - 2^x(3^x-2^x)}{x^2} \right) \)
\( = \lim_{x\to0} \left( \frac{(3^x-2^x)(3^x-2^x)}{x^2} \right) \)
\( = \lim_{x\to0} \left( \frac{(3^x-2^x)^2}{x^2} \right) \)
\( = \lim_{x\to0} \left[ \left( \frac{3^x-2^x}{x} \right)^2 \right] \)
\( = \lim_{x\to0} \left[ \left( \frac{(3^x-1) - (2^x-1)}{x} \right)^2 \right] \)
\( = \left\{ \lim_{x\to0} \left[ \frac{3^x-1}{x} \right] - \lim_{x\to0} \left[ \frac{2^x-1}{x} \right] \right\}^2 \)
\( = (\log 3 - \log 2)^2 \) \quad \( \left\{\lim_{x\to0} \left( \frac{a^x-1}{x} \right) = \log a\right\} \)
\( = \left( \log \left(\frac{3}{2}\right) \right)^2 \) ans.